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I just want to ask you if my solution is correct.

Here's the problem,

Using the Binomial Theorem, find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$.

Solution: The binomial theorem is, $$\sum_{r=0}^n{{n}\choose{r}}x^{n-r}y^r=(x+y)^n,$$Hence, by letting x = 1 and y = $\sqrt{3}$, then we have$$\sum_{r=0}^7{{7}\choose{r}}1^{7-r}{\sqrt{3}}^r=(1+\sqrt{3})^7,$$And to take the coefficient of $\sqrt{3}$ we need to set r = 1. That is,$${{7}\choose{1}}\sqrt{3}=(1+\sqrt{3})^7$$Hence, the coefficient of $\sqrt{3}$ is ${{7}\choose{1}}=7\qquad \blacksquare$

Is that correct?

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By coefficient, do you mean the 328 in: wolframalpha.com/input/… ? –  Amzoti Sep 20 '12 at 16:13

7 Answers 7

up vote 10 down vote accepted

Hint: You also get $\sqrt{3}$ terms from $r=3$, $5$, and $7$.

Remark: The following sort of uses the Binomial Theorem, but is not the intended solution. Imagine expanding $(1+\sqrt{3})^7$. Now imagine expanding $(1-\sqrt{3})^7$. Look at $$(1+\sqrt{3})^7-(1-\sqrt{3})^7.$$ If you think about the expansions, the terms that involve even powers of $\sqrt{3}$ cancel, and the ones that involve odd powers get doubled. It follows that our coefficient, which is an integer, is equal to $$\frac{(1+\sqrt{3})^7-(1-\sqrt{3})^7}{2\sqrt{3}}.$$ Divide in the calculator. To calculator accuracy, we get $328$. Since the answer is an integer, it must be $328$.

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Ok let me check that. –  Al-Ahmadgaid Asaad Sep 20 '12 at 16:10
1  
@lhf. You want a 7 exponent, not 2. –  Rick Decker Sep 20 '12 at 19:00
1  
Since $|1-\sqrt{3}|<1$, its powers get rapidly irrelevant and so the answer is $(1+\sqrt{3})^7/(2\sqrt{3})$ rounded to the nearest integer. –  lhf Sep 20 '12 at 20:42
    
@RickDecker, thanks, I've reposted the comment. –  lhf Sep 20 '12 at 20:43

You don't want only the $r=1$ term. If $r=3$ you get $$\sqrt{3}^3=\underbrace{\sqrt{3}\sqrt{3}}\sqrt{3} = 3\sqrt{3}$$ and similarly for the other terms with odd exponents.

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Wow, thank you very much... I forgot that... –  Al-Ahmadgaid Asaad Sep 20 '12 at 16:12

Let $\rm\:w = 1+\sqrt{3}.\:$ $\rm\:w^n$ has $\sqrt{3}\:$ coefficient $\rm\,b_n$ satisfying $\rm\:b_{n+2} = \color{#C00}2\:b_{n+1} + \color{#0A0}2\:b_n\:$ since $\rm\,w\,$ is a root of $\rm\:x^2 = \color{#C00}2\,x + \color{#0A0}2.\:$ Hence $\rm\,\ b_n = 0,\ 1,\ 2,\ 6,\ 16,\ 44,\ 120,\ 328,\,\ldots$ so $328$ is the answer. Details below.


Let $\rm\: w = 1\!+\!\sqrt{3},\:$ with conjugate $\rm\: \bar w = 1\!-\!\sqrt{3}.\:$ For $\rm\:v = a\!+\!b\,\sqrt{3}\:$ we get its $\sqrt{3}$ coefficient by $$\rm\: b\ =\ \frac{v - \bar v}{2\sqrt{3}}\ =\ \frac{v - \bar v}{w -\bar w}\:$$

For $\rm\:v = w^n\:$ these coefficients $\rm\:b_n\:$ satisfy a recurrence based on the minimal polynomial of $\rm\:w\:$

$$\rm w^{n+2} - \bar w^{n+2}\, = \ (\color{#C00}{w + \bar w})\ (w^{n+1} - \bar w^{n+1})\ \color{#0A0}{-\ w\bar w}\ (w^{n} - \bar w^{n})$$

$\rm Dividing\ by\, \ \ w-\bar w\, \ \ yields \ \ \ b_{n+2}\ =\ \color{#C00}2\ b_{n+1} + \color{#0A0}2\ b_n,\ \ \ b_0 = 0,\ b_1 = 1\ \ for\ \ b_n = \dfrac{w^n - \bar w^n}{w - \bar w}$

Remark $\ $ Sequeneces of the form $\rm\:\dfrac{w^n - \bar w^n}{w - \bar w}\:$ for $\rm\:w\:$ a quadratic algebraic number are known as Lucas sequences. Many special cases are well-known, e.g. the sequences of Fibonacci, Lucas, Pell, Jacobsthal, and Mersenne. These sequences arise when studying the arithmetic of quadratic number fields. The above recurrence is but one of many identities know for these and related sequences. See any of Paulo Ribenboim's "Number Records" books for many of their properties.

Note that if you express the recurrence in system (matrix) form, then you obtain a fast linear time algorithm for computing the coefficients, by repeatedly squaring matrices, just like the well-known Fibonacci case using their addition formula.

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Could you expand a little more on this, Bill? –  Pedro Tamaroff Sep 20 '12 at 17:00
    
@Peter See above. –  Bill Dubuque Sep 20 '12 at 18:08
    
If you can drop by the chat I'd like to show you something. It's about this. –  Pedro Tamaroff Sep 20 '12 at 18:33

This is not correct. You need to account for all terms in the expansion of the product where $\sqrt{3}$ is raised to an odd power. So find the sum of the terms where $r$ is odd to find your coefficient.

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Implement e Newton formula for $n=7$. Then $(1+\sqrt 3)^7=568+328\sqrt 3$. Definitely coefficient of $\sqrt 3$ is $328$.

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It is definitely –  Pedro Tamaroff Sep 20 '12 at 16:59

Write $(1+\sqrt3)^n = x_n + y_n \sqrt3$. Then $x_0=1$, $y_0=0$ and $x_{n+1}=x_n+3y_n$, $y_{n+1}=x_n+y_n$. This gives you:

n   xn  yn
0   1   0
1   1   1
2   4   2
3   10  6
4   28  16
5   76  44
6   208 120
7   568 328

You can also conclude that $x_{n+1}=y_{n+1}+2y_n$ and so $y_{n+1}=y_n+2y_{n-1}+y_n=2y_n+2y_{n-1}$, from which you can compute $y_n$ without computing $x_n$. This is Bill's answer.

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Applying the binomial theorem,

$(1+√3)^7 =1+ 7×√3 + 21×3 + 35×3√3+ 35×9 + 21×9√3 +7×27 + 1×27√3$

hence coefficient of $√3$ is $$7 + 35×3 + 21×9 + 27=328.$$

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